SGU 180 Inversions【树状数组】

题意：求逆序数

和POJ那题求逆序数的一样，不过这题离散化之后，要去一下重，然后要开到long long

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<set>
#include<queue> 
#include<algorithm>  
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int mod=1000000007;
const int maxn=80005;

int a[maxn];
int c[maxn];//树状数组

struct node{
    int x;
    int id;
} p[maxn];

int cmp(node n1,node n2){
    return n1.x < n2.x;
}

int lowbit(int x){ return x &(-x);}

__int64 sum(int x){
    __int64 ret=0;
    while(x > 0){
        ret+=c[x];x-=lowbit(x);
    }
    return ret;
}

void add(int x,int d){
    while(x < maxn){
        c[x]+=d;x+=lowbit(x);
    }
}

int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&p[i].x),p[i].id=i;
    sort(p+1,p+n+1,cmp);
    
    for(int i=1,j=0;i<=n;i++){
        if(i==1 || p[i].x != p[i-1].x) j++;
        a[p[i].id] = j;
    }
    
    memset(c,0,sizeof(c));
    __int64 ans=0;
    for(int i=1;i <=n;i++ ) {
        add(a[i],1);
        ans += i - sum(a[i]); 
    }
    printf("%I64d
",ans);
    return 0;
}